Solve this: Q 10 In a 3-digit number, unit's digit is one more than the hundred's digit and ten's - Maths - Direct and Inverse Proportions

Question

Solve this:
Q 10 In a 3-digit number, unit's digit is one more than the
hundred's digit and ten's digit is one less than the hundred's
digit If the sum of the original 3-digit number and numbers
obtained by changing the order of digits cyclically is 2664, find
the number

Varun Rawat · Accepted Answer

Let the digit at hundreds place = xNow, the digit at unit's place = x+1Now, the digit at tens place = x - 1Now, original number = 100x + 10x-1 + x+1 = 100x + 10x - 10 + x + 1 = 111x - 9Number obtained by changing the order of digits cyclically are :100x-1 + 10x+1 + x   and  100x+1 + 10x  + x - 1=111x - 90   and  111x + 99Now, according to question :     111x - 9 + 111x - 90 + 111x + 99 = 2664⇒333x = 2664⇒x = 8So, original number = 111×8-9 = 879

Solve this: Q.10. In a 3-digit number, unit's digit is one more than the hundred's digit and ten's digit is one less than the hundred's digit. If the sum of the original 3-digit number and numbers obtained by changing the order of digits cyclically is 2664, find the number.

Solve this:
Q.10. In a 3-digit number, unit's digit is one more than the hundred's digit and ten's digit is one less than the hundred's digit. If the sum of the original 3-digit number and numbers obtained by changing the order of digits cyclically is 2664, find the number.